00:01
Hi, so to solve for this, we need to determine first the limiting reactant.
00:05
We'll convert the given mass of reactants to moles.
00:08
Let's start with acetylene c2h2.
00:10
If we have 26 grams of this compound, convert to moles by dividing the molar mass of c2h2.
00:17
That's 26 .04 grams and then we have here moles, so that means you could cancel grams.
00:24
And this will give us 0 .998.
00:28
This is moles of acetylene.
00:31
And then we have oxygen gas, number of moles of o2.
00:34
If we have 320 grams of o2, convert to moles by dividing the molar mass, which is 32 grams per mole of oxygen gas.
00:43
So we could cancel grams and this will give us 10 moles of o2...